Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.StringTokenizer;

public class P816C
{
	static ArrayList<String> list = new ArrayList<>();
	public static void main(String[] args)
	{
		FastScanner scan = new FastScanner();
		int n = scan.nextInt();
		int m = scan.nextInt();
		int[][] grid = new int[n][m];
		for (int i = 0; i < n; i++)
			for (int j = 0; j < m; j++)
				grid[i][j] = scan.nextInt();
		if (n <= m)
		{
			decRows(n, m, grid);
			decCols(n, m, grid);
		}
		else
		{
			decCols(n, m, grid);
			decRows(n, m, grid);
		}
		for (int i = 0; i < n; i++)
		{
			for (int j = 0; j < m; j++)
			{
				if (grid[i][j] != 0)
				{
					System.out.println(-1);
					return;
				}
			}
		}
		PrintWriter pw = new PrintWriter(System.out);
		pw.println(list.size());
		for (int i = 0; i < list.size(); i++)
			pw.println(list.get(i));
		pw.flush();
	}

	private static void decCols(int n, int m, int[][] grid)
	{
		for (int i = 0; i < m; i++)
		{
			int min = Integer.MAX_VALUE;
			for (int j = 0; j < n; j++)
				min = Math.min(min, grid[j][i]);
			for (int j = 0; j < n; j++)
				grid[j][i] -= min;
			for (int j = 0; j < min; j++)
				list.add("col " + (i+1));
		}
	}

	private static void decRows(int n, int m, int[][] grid)
	{
		for (int i = 0; i < n; i++)
		{
			int min = Integer.MAX_VALUE;
			for (int j = 0; j < m; j++)
				min = Math.min(min, grid[i][j]);
			for (int j = 0; j < m; j++)
				grid[i][j] -= min;
			for (int j = 0; j < min; j++)
				list.add("row " + (i+1));
		}
	}
	
	static class FastScanner
	{
		BufferedReader br;
		StringTokenizer st;

		public FastScanner()
		{
			try
			{
				br = new BufferedReader(new InputStreamReader(System.in));
				st = new StringTokenizer(br.readLine());
			} catch (Exception e)
			{
				e.printStackTrace();
			}
		}

		public String next()
		{
			if (st.hasMoreTokens())
				return st.nextToken();
			try
			{
				st = new StringTokenizer(br.readLine());
			} catch (Exception e)
			{
				e.printStackTrace();
			}
			return st.nextToken();
		}

		public int nextInt()
		{
			return Integer.parseInt(next());
		}

		public long nextLong()
		{
			return Long.parseLong(next());
		}

		public String nextLine()
		{
			String line = "";
			try
			{
				line = br.readLine();
			} catch (Exception e)
			{
				e.printStackTrace();
			}
			return line;
		}
	}
}